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molecular speed, the gas in the left half of the box will be cooled and the gas in the right half will be heated. A number of devices have been proposed thatt according to their authors, could play the role of such a demon, but all of them failed to work because of certain unanticipated effects found in them. However, where physics faiis, mathematics may succeed. I want to suggest a device, free from demons and other evil spirits, that can be built even in a high school workshop. I'11 present calcuiations that wili clearly show that this device can transfer heat from one body to another if they were heated equally at ihe outset. Before going into a description of my "perpetual enging" you'llhave to swallow a certain amount of math so that you can convince yourself that the reasoning and constructions to follow are correct. We'llbe dealing with the ellipse and its properties. By definitior', an ellipse is the cuwe formed by alJ the points in the plane such that the sum of their distances to two fixed points F, and F, is constant. Each of the points Fr and F, is called afocusi the constant sum oI the distances is usually denoted by 2a. This property is used by gardeners when they want to make oval flower beds. They drive a pair of sticks into the ground (at the foci), tie the ends of a rope to them, then take another stick with a sharp end and stretch the rope taut with this stick. If the stick is moved so that the rope remains taut, its sharp end traces an ellipse. The size and shape of the ellipse depend on the length of the rope and the distance between the foci. You can verify this on a sheet of paper using two pins and a pencil instead of sticks {{ig.2l. After you draw several ellipses you'll see that any ellipse is a closed convex curve that has a center of symmetry and two symmetry axes-the line FrF, and the Perpendicular bisector of the segmentFf 2. Also, it's easy to see that the sum of the distances to the foci is less than 2a lor points inside the ellipse and greater than2a outside it. This information suffices to prove an important and not so obvious property of the ellipse: the segments that join the foci of an ellipse to an arbitrary point M on it make equal angles with the line that touches the ellipse at M. Comparing this property to the law of reflection of light-the angle of incidence equals the angle of reflection-we come up with the following formulation: a tay of light issuing from a focus of an eliipse after reflection from it hits the other focus. This is the "optical property" of the ellipse. It can be observed in nature: there are caves with eilipsoidal domes where you can find two spots/ far enough from each other, such that the voice of a person standing at one of these spots is heard at the other spot as if the speaker were just inches away. And some palaces and castles have halls intentionally designed to produce this effect. Since the optical property of the ellipse plays a major role in what follows, I'11give its proof-it's rather short and simple. Let l be the tangent to an ellipse at its point M lfig.3) and let cr and p be the angles between the linel and segments F, M and MF rlF, and F, are the foci). Reflect F, about I into F1', ioin F r'F2, and find the intersection pointNof this segment withl. SuPpose N * M; then N lies outside the ellipse (actually, the entire line 1, except its point M, lies outside the ellipse, because any ellipse is a convex curve). Therefore FrN + NF, > 2a : FrM + MFr. But by the construction, FrN = Fr'N a<strong>nL</strong>dFrM = Ft'M, so we get Fr'N + NFz> Fr'M + MF, ot Fr'M + MFr. Fr'F, whtch contradicts the Triangle Inequality for triatgle Fr'FrM. Therefore, points N andM must coincide, so FrMFr'is a straight line-that is, cr = P. Now let's bring our project to fruition. Take a sheet of good drawing paper, mark points F, and F, on it, and &aw two ellipses with the foci at these points, using a longer string the first time and a shorter one the second time. Draw the Perpendicular bisector to the segment FrErand erase a part of what we've drawn so as to obtain a "mushroom" like the one in figure 4. Figure 4 Nowrollyour "blueprint" up, Put it in a cardboard tube, and go to the nearest metalworking shop. Ask that atin sheil be made in the shape obtained by rotating the curve in the blueprint about its symmetry axis f rF, (fiS. 5). The inside of the shell must be covered with a reflective coating. When the thing is readY, take it home. Now you are the Figure 2 Figure 3 Figure 5 JlJr.Y/AU0USI 1S04
Figure 6 owner of a perpetual engine! You don't believe me? We1l, I'll prove it mathematically. Place two equally heated bodies at points F, and Fr. Notice that any ray lof light or heat) issuing from F, must arrive atFzffig.6l. Aray propagated along the same trajectory from F, will arrive back at F, of course. But look at a ray aimed frorn F, at the vertical barrier. If there were no partition, this ray would rebound from the erased part of the ellipse and hit Fr. But now it will be reflected at some point R of the partition in the direction it would have had if it had propagated from F, to R (fig. 6). Therefore, it will come back to Fr! And, clearly,leaving point F, along the same trajectory in the opposite direction, we again come back to Fr. So there's a considerable number of rays that make "light loops." In the cross section of our "engine" shown in figure 5, these are aLL the rays within the angle AFrB and the symmetric angle A',FzB'.. So, the body at F, will be heated, while the body at F, willbe cooled. We can even compute the eventual temperatures of the bodies after the process stabilizes. Suppose the initial temperatures were 7o K (or "kelvins," named after the English physicist W. Thomson mentioned above, on whom the title of Lord Kelvin was con{erred for his outstanding scientific achievements). It's known that the rate of energy loss by radiation (the luminosity) is proportional to the fourth power o{ the temperature (in kelvins) of the radiating body. Therefore, if the temperature at Figure 7 point F, is 7r, and at Frit's Tr, then the luminosity at F, and F, canbe written as k( andkTf , respectively. A11 the rays from F, after one reflection arrive at F,b:ut only some of the rays from F, arrive at Fr. How many? Consider a small sphere centered atFr. The rays that return to F, cut a ring out of this sphere lfi4. Tl- If its area is A, and the total area of the sphere is A, than the amount of radiation that comes back to F, equals ArkTtlA, and the portion that arrives at F, is (A - A)kftlA. In the steady state, the energy radiated from point F, equals the energy that arrives at this point in the same time interval. That is, (o- +)U = krl. \t/A On the other hand, by the law of conservation of energy, the loss of energy atF, equals the gain of energy atFz-thatis, C(7, -Tol: ClTo-721, where C is the heat capacity of either body,(we assume that the bodies are absolutely alike). The last equation gives 7, + Tr:2T0, so using the previous relationship and letting b = l(A - Ar) I A| la, andnoting that b < 1, we finally get T, ' =T&.To, r+b 2T^ "= ri"o' That's it! You can check the reasoning from the very beginning and be satisfied that it's perfectly correct. So, did we really refute the second law of thermodynamics? Unfortunately (or fortunately), no. We've made a mistake. But where? Think about it yourself, and then compare your answer with what's written below. I have a feeling that not every reader will be able to find the correct answer. You were on the wrong track if you tried to find an error in the physics-for instance, in the fact that we ignored convection. Why? We can cteate a vacuum inside the shell. But what else is there in our little "engine" except math and physics? The error is hiding at the border, so to speak, between these two sciencesin the transition from the physical process to its mathematical model. Recall that we dealt with two bodies placed at the foci. Thus, we neglected the sizes of the bodies. This is common in physics reasoning, and the phrase "a body is 1ocated at point M" doesn't cause anyone to protest. In many situations, this disregard of the sizes of the bodies is justifiable. For instance the motion of a body under given forces applied to its center of mass doesn't depend on its size and shape. Or, as Newton showed, a body consisting of a number of concentric homogeneous spherical layers attracts other bodies as the same mass concentrated at the center of this body. So to reptrace a body with a point is indeed a customary operation in physics. But in our case this operation leads to an error. Let's see whether our reasoning remains valid if we consider balls of nonzero radius centered at the foci. Consider three rays reflected from the ellipse at the same point M in its left half (fig. B): a ray issuing from the foci F, and two rays, AM arrd BM, with endpoints at a distancer fr.or;-i,Fr. M IB Figure B Y', OUANTUllll/IIATUBE
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